phân tích đa thức thành nhân tử
3x(x 1)^2-5x^2(x 1) 7(x 1)
Những câu hỏi liên quan
Phan tich da thuc thanh nhan tu x^2*[(x^2+1/x^2)+6*(x-1/x)+7]
Phan tich da da thuc thanh nhan phan tu
(x^2+x+1)(x^2+x+2)-12
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)
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phan tich da thuc thanh nhan tu
3x(x+1)2-5x2(x+1)+7(x+1)
phan tich da thuc thanh nhan tu
a, x^7+x^2+1
b, x^7+x^5+1
a) \(x^7+x^2+1\)
\(=x^7-x+x+x^2+1\)
\(=\left(x^7-x\right)+\left(x^2+x+1\right)\)
\(=x\left(x^6-1\right)+\left(x^2+x+1\right)\)
\(=x\left(x^3+1\right)\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^4+x\right)\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^5-x^4+x^2-x\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^5-x^4+x^2-x+1\right)\left(x^2+x+1\right)\)
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b) \(x^7+x^5+1\)
\(=x^7+x^6+x^5-x^6+1\)
\(=\left(x^7+x^6+x^5\right)-\left(x^6-1\right)\)
\(=x^5\left(x^2+x+1\right)-\left(x^3+1\right)\left(x^3-1\right)\)
\(=x^5\left(x^2+x+1\right)-\left(x^4-x^3+x-1\right)\left(x^2+x+1\right)\)
\(=\left(x^5-x^4+x^3-x^2+1\right)\left(x^2+x+1\right)\)
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\(x^7+x^2+1\)
\(=\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)
\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
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phan tich da thuc sau thanh nhan tu (x-1)(x-3)(x-5)(x-7)-20
\(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20=\left[\left(x-1\right)\left(x-7\right)\right].\left[\left(x-3\right)\left(x-5\right)\right]-20\)
\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)
Đặt \(x^2-8x+11=t\) \(\Rightarrow\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20=\left(t-4\right)\left(t+4\right)-20=t^2-16-20=t^2-36=\left(t-6\right)\left(t+6\right)\)\(\Rightarrow\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20=\left(x^2-8x+11-6\right)\left(x^2-8x+11+6\right)=\left(x^2-8x+17\right)\left(x^2-8x+5\right)\)
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phan tich da thuc thanh nhan tu x4+x7+1
Phan tich da thuc thanh nhan tu (1+2x)(1-2x)-x(x+2)(x-2)
\(\left(1+2x\right).\left(1-2x\right)-x.\left(x+2\right).\left(x-2\right)\))
\(=1-\left(2x\right)^2-x.x^2-2^2\)
\(=1-4x^2-x^3-4\)
Ko bt có đúng ko nữa
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( 1 + 2x ) ( 1 - 2x ) - x ( x + 2 ) ( x - 2 )
= 1 - 4x2 - x ( x2 - 4 )
= 1 - 4x2 - x3 + 4x
= - ( x3 + 4x2 - 4x - 1 )
= - ( x3 - x2 + 5x2 - 5x + x - 1 )
= - [ x2 ( x - 1 ) + 5x ( x - 1 ) + ( x - 1 ) ]
= - ( x - 1 ) ( x2 + 5x + 1 )
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Phan tich da thuc thanh nhan tu
x^7+x^5+1
Phan tich da thuc thanh nhan tu
x(x+1)(x+2) + 2*(x+3)+1
phan tich da thuc thanh nhan tu: x(x+2)(x^2+2x+2)+1
x(x+2)(x^2+2x+2)+1 = (x^2+2x)(x^2+2x+1)+1
Đặt x^2+2x+1=y ta được:
(y-)(y+1)+1=y^2-1+1=y^2
= (x^2+2x+1)^2
= ( x + 1 )^4
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